---
title: "BCC, FCC and HCP crystal structures"
subject: "Material Science"
chapter: 2
key: "semester_3/material_science:bcc-fcc-hcp-structures"
source: https://www.bctengineer.com/q/semester_3/material_science/bcc-fcc-hcp-structures
site: BCT Engineering
---

# BCC, FCC and HCP crystal structures

Material Science · Chapter 2: Atomic Structure, arrangement of atoms

- Asked in: year not given, 6 marks
- Video: [Materials Science Explained: Crystal Structures, FCC BCC HCP & Tin Disease Mystery](https://www.youtube.com/watch?v=PdFPIcSdxBk) — Ibham Veza
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/material_science/bcc-fcc-hcp-structures
- Whole chapter: https://www.bctengineer.com/md/semester_3/material_science/chapter-2

## Question

Describe the BCC, FCC and HCP crystal structures. For BCC and FCC, determine the number of atoms per unit cell, the relation between atomic radius and lattice constant, the coordination number and the atomic packing factor.

## Short answer

BCC: 2 atoms/cell, R = (√3/4)a, CN = 8, APF = 0.68. FCC: 4 atoms/cell, R = (√2/4)a, CN = 12, APF = 0.74. HCP: 6 atoms/cell, ideal c/a = 1.633, CN = 12, APF = 0.74. FCC and HCP are close-packed, differing in stacking sequence (ABC vs AB).

## Answer

**Atomic packing factor (APF)** is the fraction of the unit-cell volume occupied by atoms:

$$
\text{APF} = \frac{n \times \frac{4}{3}\pi R^3}{a^3}
$$

#### BCC (e.g., $\alpha$-Fe, Cr, W)
Atoms at the 8 corners and one at the body centre. Atoms touch along the body diagonal.

- Atoms per cell: $8 \times \tfrac{1}{8} + 1 = 2$
- Body diagonal $= \sqrt{3}\,a = 4R$, so $R = \dfrac{\sqrt{3}\,a}{4}$
- Coordination number = 8

$$
\text{APF} = \frac{2 \times \frac{4}{3}\pi\left(\frac{\sqrt{3}a}{4}\right)^3}{a^3} = \frac{\sqrt{3}\,\pi}{8} = 0.68
$$

#### FCC (e.g., $\gamma$-Fe, Cu, Al, Ni)
Atoms at the 8 corners and the centre of each of the 6 faces. Atoms touch along the face diagonal.

- Atoms per cell: $8 \times \tfrac{1}{8} + 6 \times \tfrac{1}{2} = 4$
- Face diagonal $= \sqrt{2}\,a = 4R$, so $R = \dfrac{\sqrt{2}\,a}{4}$
- Coordination number = 12

$$
\text{APF} = \frac{4 \times \frac{4}{3}\pi\left(\frac{\sqrt{2}a}{4}\right)^3}{a^3} = \frac{\pi}{3\sqrt{2}} = 0.74
$$

#### HCP (e.g., Zn, Mg, Ti, Co)
Two hexagonal layers (top and bottom) with a central atom each, plus a triangular layer of 3 atoms midway between them. Atoms per cell = $12 \times \tfrac{1}{6} + 2 \times \tfrac{1}{2} + 3 = 6$. Ideal axial ratio $c/a = 1.633$, coordination number 12, APF = 0.74.

| Structure | Atoms/cell | CN | APF |
|---|---|---|---|
| BCC | 2 | 8 | 0.68 |
| FCC | 4 | 12 | 0.74 |
| HCP | 6 | 12 | 0.74 |

FCC and HCP are both close-packed, differing in stacking sequence (ABCABC for FCC, ABABAB for HCP).

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Along which direction do atoms touch in BCC and what is R?
   A. Face diagonal; R = sqrt(2)a/4
   B. Body diagonal; R = sqrt(3)a/4
   C. Cube edge; R = a/2
   D. Body diagonal; R = sqrt(2)a/4
   Answer: B. Body diagonal = sqrt(3)a = 4R in BCC. (Key point: BCC contact direction)

2. How many atoms per unit cell in BCC and FCC?
   A. 2 and 4
   B. 1 and 4
   C. 2 and 6
   D. 4 and 4
   Answer: A. BCC: 8x1/8 + 1 = 2; FCC: 8x1/8 + 6x1/2 = 4. (Key point: Atoms per cell)

3. Which gives the correct APF and coordination number for FCC?
   A. 0.68, CN 8
   B. 0.52, CN 6
   C. 0.74, CN 12
   D. 0.74, CN 8
   Answer: C. FCC is close packed: APF 0.74, CN 12. (Key point: Packing factor)

4. How do FCC and HCP differ?
   A. Stacking sequence: ABCABC vs ABABAB
   B. APF: 0.74 vs 0.68
   C. Coordination number: 12 vs 8
   D. FCC is not close packed
   Answer: A. Both are close packed with same APF and CN, differing in stacking sequence. (Key point: FCC vs HCP)

## Related in Material Science

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- [Critical nucleus radius for copper](https://www.bctengineer.com/question.md?key=semester_3%2Fmaterial_science%3Acritical-nucleus-radius-copper) — Material Science, chapter 5

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